2012 #3 Flowers

Flowers

Time Limit:2000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

Submit Status Practice HDU 4325

Description

As is known to all, the blooming time and duration varies between different kinds of flowers. Now there is a garden planted full of flowers. The gardener wants to know how many flowers will bloom in the garden in a specific time. But there are too many flowers in the garden, so he wants you to help him.

 

Input

The first line contains a single integer t (1 <= t <= 10), the number of test cases. 
For each case, the first line contains two integer N and M, where N (1 <= N <= 10^5) is the number of flowers, and M (1 <= M <= 10^5) is the query times. 
In the next N lines, each line contains two integer S _i and T _i (1 <= S _i <= T _i <= 10^9), means i-th flower will be blooming at time [S _i, T _i].
In the next M lines, each line contains an integer T _i, means the time of i-th query. 

 

Output

For each case, output the case number as shown and then print M lines. Each line contains an integer, meaning the number of blooming flowers. 
Sample outputs are available for more details. 

 

Sample Input

2
1 1
5 10
4
2 3
1 4
4 8
1
4
6

 

Sample Output

Case #1:
0
Case #2:
1
2
1

 #include <stdio.h>
 #include <string.h>
 #include <algorithm>
 using namespace std;
 int main()
 {
     int T;
     int i,j;
     int n,m,ca=;
     int l[],r[];
     scanf("%d",&T);
     while(T--)
     {
         scanf("%d %d",&n,&m);
         for(i=;i<=n;i++)
             scanf("%d %d",&l[i],&r[i]);
         sort(l+,l+n+);
         sort(r+,r+n+);
         printf("Case #%d:\n",ca);ca++;
         for(i=;i<=m;i++)
         {
             int x,y,z;
             scanf("%d",&z);
             x=upper_bound(l+,l+n+,z)-(l+);
             y=upper_bound(r+,r+n+,z-)-(r+);
             printf("%d\n",x-y);
         }
     }
     return ;
 }